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<>model:<BR>!合同拖期惩罚费用最小;<BR>sets:<BR> Contract/1..18/:W,C,ys,Dely,t,A; !18个合同,w交货量,C交货期,A惩罚系数;<BR> Time/1..6/;!6个计划半旬; <BR> CTX(Contract,Time):x;!x表示合同i在t内加工,;<BR>endsets<BR>data:<BR> A=3.472 3.37 2.71 5.65 3.078 3.756 2.563 3.589 6.159 3.9526 2.37 2.21 5.65 3.178 2.756 2.863 3.189 4.159; <BR> W=307.6 316 289.7 300 150 146 200 170 150 150 200 273.51 595.2 464 500 500 137.14 246.87;<BR> c=1 2 3 1 2 2 3 2 1 2 3 2 3 4 1 2 3 2;<BR> ys=3743.610548 3262.520165 2358.480326 2442.333786 812.9163374 2617.204033 2527.64613 3047.429354 2688.908253 2688.908253 2434.077079 1200.043876 2592.052259 3946.137491 4252.303331 4252.303331 479.16145 3120;<BR>enddata <BR>!目标;<BR><a href="mailtmin=@sum(Contract(i" target="_blank" >min=@sum(Contract(i</A>):<BR> A(i)*W(i)*dely(i));</P>
<>@for(Contract(i):<BR> @sum(CTX(i,j):X(i,j))=1);</P>
<>@for(Contract(i):<BR> dely(i)=@smax(t(i)-C(i),0)); !求拖期;</P>
<P>!生产能力约束;<BR>@for(time(j):<BR> @sum(CTX(i,j):<BR> x(i,j)*ys(i))<8000); </P>
<P>@for(ctx(i,j): t(i)=@if(X(i,j)#eq#1,k,t(i)));</P>
<P><BR>@for(CTX(i,j)bin(X));!x为0-1整数变量;<BR>end</P>
<P>怎样才能获得一个优解</P> |
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